Solve the given differential equation. (d y)/(d x)=(x √(x^2+y^2)+y^2)/(x y), x>0

Name: Problem 23, Chapter 1: Differential Equations and Linear Algebra
Uploaded: 2020-06-16T13:14:40-08:00
Duration: 7 min 9 s
Channel: Samuel Hannah
Description: Solve the given differential equation. (d y)/(d x)=(x √(x^2+y^2)+y^2)/(x y), x>0

Solve the given differential equation. (d y)/(d x)=(x...

Key Concepts

Homogeneous Differential Equations

A differential equation is termed homogeneous if the functions involved can be expressed as a function of the ratio y/x (or a similar variable ratio). Recognizing the equation as homogeneous enables the transformation of variables, such as substituting y = vx, which simplifies the equation by reducing it to a separable form.

Ordinary Differential Equations

This concept involves equations containing functions and their derivatives. It is fundamental in understanding how quantities change and relates rates of change to the functions themselves. The given problem is a first-order differential equation, which means it involves only the first derivative of the unknown function with respect to the independent variable.

Variable Substitution

Variable substitution, such as setting y = vx, is a key strategy in solving homogeneous differential equations. This substitution transforms the differential equation into one where the new variable v and its derivative can be separated from the independent variable. This method significantly simplifies the integration process.

Separation of Variables

After an appropriate substitution, the differential equation often becomes separable. This means that the variables can be isolated on opposite sides of the equation, allowing for direct integration with respect to each variable. Separation of variables is a widely-used technique for solving first-order differential equations.

Integration Techniques

Once the equation is separated, integration techniques are applied to both sides to find the general solution. This involves integrating functions that may require standard methods like substitution, integration of rational functions, or recognition of standard integral forms to arrive at the implicit or explicit solution of the differential equation.

Transcript

00:01 So for this problem, we have a first order differential equation, which i have rewritten in the top left of the screen.

00:06 Now, we would like to rewrite this equation in terms of homogeneous variables.

00:14 And to do that in a more explicit form, we want to get everything in terms of y over x.

00:19 We will start this by looking at the square root.

00:23 So if we factor out x squared underneath the square root, we get that the square root is equal to the square root.

00:30 Of x squared multiplied by the square root of 1 plus y squared over x squared, and then this is equal to x times the square root of 1 plus y squared over x squared.

00:46 Now we take the positive square root of x, because by the question, x is always positive.

00:53 So the right hand side of our differential equation now reads x squared, times the square root of 1 plus y squared over x squared, plus y squared all over x y.

01:10 And we can do one final rewriting by factoring out x squared in the numerator and the denominator, which gives us that y d x is equal to the square root of 1 plus y squared over x squared plus y squared over x squared plus y squared over x squared all over y over x this we see that we now have the differential equation in a form of being only dependent on y over x so it is in fact homogeneous we can then make the variable transformation of v equals y over x or y equals vx equivalently using the second formulation we can see that that d -y -d -x is equal to v plus x, evdx using the product rule.

02:21 If we substitute this into the left -hand side of our equation, and for the right -hand side, we just substitute in v, we get the differential equation being v -plus -x -e -v -d -x is equal to the square root of 1 plus v squared, plus v squared, over v.

02:48 We can make an additional simplification by subtracting v from both sides, but on the right -hand side, notice that if we were to split the fraction into its component forms, the second component, v squared over v, is simply v itself, and v minus v vanishes.

03:08 So the differential equation is simply x, vv the x is equal to the square root 1 plus, v squared over v.

03:22 This is a separable differential equation, so we will separate the variables.

03:29 Bring v to the left hand side, you get that v over the square root of 1 plus v squared, v is equal to 1 over x the x...

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